# Radiation Pressure in Space

1. May 23, 2012

### tempneff

1. The problem statement, all variables and given/known data
A motionless spacecraft powered by a “solar sail” at a distance of 1.20 AU
from the sun has a perfectly reflective sheet of lightweight material that allows the spacecraft to accelerate by using the radiation pressure from the sun. If a certain sail is in the shape of a square of side 800 m, its velocity is 38.5 km/s when it reached a distance of 3.50 AU from the sun. The Force at 1.2 AU is 4.18 N and at 3.54 AU is 0.492.
Answer range: 3.0 to 3.4 years

2. Relevant equations

$$a= \frac{v_f^2-v_0^2}{2(x_f-x_0}$$
$$t=\frac{v_f-v_0}{a}$$
$$1 AU \approx 1.5\times 10^8 km$$
$$1 year \approx 3.16 \times 10^7 s$$

3. The attempt at a solution

I tried these kinematics equations and can't see his answer range, so I think I am missing
something.
$$t=\frac{2(x_f-x_0)}{v_f}=1.79\times 10^7 s\neq 3 years-ish$$
I assume I can't use constant acceleration, no?

2. May 23, 2012

### Staff: Mentor

since the newtons change then by definition the acceleration changes too. What if you average the initial and final accelerations computed from the force of radiation/ mass of ship at start and end. that should give a lower acceleration and longer time range.

3. May 23, 2012

### tempneff

I'm not sure it's that simple. I tries it and am still of about 1/5 of where I should be. At 1.2 AU the craft is motionless, why is this if it has a force of 0.492 N shouldn't it be accelerating?

Correction: it's motionless relative to the sun

Last edited: May 23, 2012
4. May 23, 2012

### tempneff

So I tried to get an acceleration average using

$$da=\frac{dF}{m}$$

So

$$a=\frac{1}{m} \int_{F_1}^{F_2}dF$$

These numbers are not working out either

5. May 23, 2012

### Staff: Mentor

okay so it starts motionless (vi=0) at 1.2AU the solar radiation applies a force of 4.18N to the 800m^2 sail so it will begin to pick up speed as it travels away from the sun and at the same time the solar radiation force will decrease to .492N at 3.54AU.

If you notice that light like gravity follows an inverse square law: F=C / (r^2) to find C plugin the F value and
convert the r value to meters. When I do it with the problem numbers I get C=1.33E17 in the 4.18 case and 1.37E17 in the 0.492 case

then this problem is similar to the one where a spacecraft is stationary out in space and then falls to Earth and you want to know how long it takes to fall a given distance remembering that as the spacecraft falls the Earth's pull on it gets stronger and stronger.

Also the problem now reminds me of one I worked in classical physics long ago where we were asked to validate that an object falling toward earth from a great distance would traverse first half the dist in something like 3/5 of the time (I cant remember the exact ratio). We just couldn't connect the time to the distance fallen until the prof mentioned using Kepler's 3rd law on orbits and collapsing the orbit minor axis to zero. You might check wikipedia searching kepler law or free fall time for more details.

6. May 24, 2012

### Staff: Mentor

had another thought don't you need to factor in the pull of the sun on the spacecraft? that would reduce its acceleration by a bit.

7. May 25, 2012

### tempneff

I have a function for the velocity which I am going to integrate for the time. The integral is a strange one though. I am going to post my work in the math section for help.

8. May 27, 2012

### tempneff

Okay this is the function I have for velocity and I think I have the integration....

$$\Delta K$$=Work due to gravity (GMm/radius) + work due to radiation pressure(I'll call it p/radius)

$$\frac{1}{2}v^2=\frac{GMm}{r_2}-\frac{GMm}{r_1}+\frac{p}{r_2}-\frac{p}{r_1}$$
factor..
$$\frac{1}{2}v^2=\bigg( \frac{p}{m}-GM\bigg) \bigg( \frac{1}{r_1}-\frac{1}{r_2} \bigg)$$
so
$$v=\sqrt{2\bigg( \frac{p}{m}-GM\bigg) \bigg( \frac{1}{r_1}-\frac{1}{r_2} \bigg)}$$
Let
$$2\bigg( \frac{p}{m}-GM\bigg)=b$$
so
$$v=\sqrt{b \bigg( \frac{1}{r_1}-\frac{1}{r_2} \bigg)}$$
or
$$v=\sqrt{b \bigg( \frac{r_2-r_1}{r_1r_2} \bigg)}$$
then
$$\frac{1}{v}=\sqrt{\frac{r_1r_2}{b(r_2-r_1)}}$$

$$v=\frac{dr}{dt}$$
and
$$dt=\frac{dr}{v}$$
then
$$t=\int \frac{1}{v}dr$$
and.....
$$t=\int \sqrt{ \bigg( \frac{r_1r_2} {b(r_2-r_1)}\bigg)}dr$$

I think so.....then for easier typing let $r_1=a$ and $r_2=x$
$$t=\int \sqrt{\bigg( \frac{ax} {b(x-a)}\bigg)}dx$$
now if we let
$$x=a\sec^2 \theta$$then$$dx=2a\sec^2 \theta \tan \theta d \theta$$
and$$t=\frac{1}{\sqrt{b}}\int \sqrt{\frac{a^2 \sec^2 \theta}{a \sec^2 \theta - a}}2a \sec^2 \theta \tan \theta d \theta$$
$$t=\frac{1}{\sqrt{b}}\int \sqrt{\frac{a^2 \sec^2 \theta}{a \tan^2 \theta}}2a \sec^2 \theta \tan \theta d \theta$$
$$t=\frac{1}{\sqrt{b}}\int \frac{a\sec \theta}{\sqrt{a} \tan \theta}2a \sec^2 \theta \tan \theta d \theta$$
$$t=\frac{2a^2}{\sqrt{ab}}\int \sec^3 \theta d \theta$$
By integral table
$$\int sec^3 \theta d \theta = \frac{1}{2} \bigg( \sec \theta \tan \theta +\ln |\sec \theta + \tan \theta| \bigg)+C$$
If $$x=a \sec^2 \theta$$
then$$\sec \theta = \sqrt{\frac{x}{a}}$$
and
$$\tan \theta = \sqrt{x-a}$$
so
$$t=\frac{a^2}{\sqrt{ab}}\bigg( \sqrt{\frac{x(x-a)}{a}}+\ln \bigg|\sqrt{\frac{x}{a}}+\sqrt{x-a}\bigg| \bigg) +C$$

Last edited: May 27, 2012
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